Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $q = \dfrac{-7y - 49}{y + 2} \div \dfrac{y^2 + 13y + 42}{8y^2 + 48y} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-7y - 49}{y + 2} \times \dfrac{8y^2 + 48y}{y^2 + 13y + 42} $ First factor the quadratic. $q = \dfrac{-7y - 49}{y + 2} \times \dfrac{8y^2 + 48y}{(y + 7)(y + 6)} $ Then factor out any other terms. $q = \dfrac{-7(y + 7)}{y + 2} \times \dfrac{8y(y + 6)}{(y + 7)(y + 6)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ -7(y + 7) \times 8y(y + 6) } { (y + 2) \times (y + 7)(y + 6) } $ $q = \dfrac{ -56y(y + 7)(y + 6)}{ (y + 2)(y + 7)(y + 6)} $ Notice that $(y + 6)$ and $(y + 7)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -56y\cancel{(y + 7)}(y + 6)}{ (y + 2)\cancel{(y + 7)}(y + 6)} $ We are dividing by $y + 7$ , so $y + 7 \neq 0$ Therefore, $y \neq -7$ $q = \dfrac{ -56y\cancel{(y + 7)}\cancel{(y + 6)}}{ (y + 2)\cancel{(y + 7)}\cancel{(y + 6)}} $ We are dividing by $y + 6$ , so $y + 6 \neq 0$ Therefore, $y \neq -6$ $q = \dfrac{-56y}{y + 2} ; \space y \neq -7 ; \space y \neq -6 $